Problem: Find all $a,$ $0^\circ < a < 360^\circ,$ such that $\cos a,$ $\cos 2a,$ and $\cos 3a$ form an arithmetic sequence, in that order.  Enter the solutions, separated by commas, in degrees.
Solution: We want $a$ to satisfy
\[\cos a + \cos 3a = 2 \cos 2a.\]By the double-angle and triple-angle formula, this becomes
\[\cos a + (4 \cos^3 a - 3 \cos a) = 2 \cdot (2 \cos^2 a - 1).\]This simplifies to
\[4 \cos^3 a - 4 \cos^2 a - 2 \cos a + 2 = 0,\]which factors as $2 (\cos a - 1)(2 \cos^2 a - 1) = 0.$  Hence, $\cos a = 1,$ $\cos a = \frac{1}{\sqrt{2}},$ or $\cos a = -\frac{1}{\sqrt{2}}.$

The equation $\cos a = 1$ has no solutions for $0^\circ < a < 360^\circ.$

The equation $\cos a = \frac{1}{\sqrt{2}}$ has solutions $45^\circ$ and $315^\circ.$

The equation $\cos a = -\frac{1}{\sqrt{2}}$ has solutions $135^\circ$ and $225^\circ.$

Thus, the solutions are $\boxed{45^\circ, 135^\circ, 225^\circ, 315^\circ}.$